Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M

51.

The sum of two positive numbers is 240 and their HCF is 15. Find the number of pairs of numbers satisfying the given condition.

• (A) 5
• (B) 2
• (C) 8
• (D) 4

Show Answer

 15(x + y) = 240 x + y = 16 1 + 15 (Right) 2 + 14 (Wrong) 3 + 13 (Right) 4 + 12 (Wrong) 5 + 11 (Right) 6 + 10 (Wrong) 7 + 9 (Right) 8 + 8 Total pairs = 4

52.

Which of the following is a pair of co-primes ?

• (A) (16, 12)
• (B) (18, 25)
• (C) (21, 35)
• (D) (23, 92)

Show Answer

 H.C.F. of 18 and 25 is 1 So, they are co-primes

53.

The smallest square number divisible by 10, 16 and 24 is ?

• (A) 900
• (B) 1600
• (C) 2500
• (D) 3600

Show Answer

 LCM (10, 16, 24) = 5 × 2 × 8 × 3 = 240 ⇒ for square number split the LCM into its factors = 5 × 2 × 2 × 2 × 2 × 3 = 5 × 5 × 2 × 2 × 2 × 2 × 3 × 3 = 3600

54.

A fraction becomes $$\frac{1}{6}$$ when 4 is subtracted from its numerator and 1 is added to its denominator. If 2 and 1 are respectively added to its numerator and the denominator, it becomes $$\frac{1}{3}$$. Then, the LCM of the numerator and denominator of the said fraction, must be = ?

• (A) 14
• (B) 350
• (C) 5
• (D) 70

Show Answer

 $$\eqalign{ & {\text{Let fraction is }}\frac{x}{y} \cr & \therefore \frac{{x - 4}}{{y + 1}} = \frac{1}{6}({\text{given}}) \cr & \Rightarrow {\text{cross multiply the equation }} \cr & \Rightarrow {\text{6}}x - 24 = y + 1 \cr & 6x - y - 25 = 0......({\text{i}}) \cr & {\text{Again,}} \cr & \frac{{x + 2}}{{y + 1}} = \frac{1}{3}({\text{given}}) \cr & \Rightarrow 3x + 6 = y + 1 \cr & 3x - y + 5 = 0.......({\text{ii}}) \cr & {\text{From equation (i) and (ii)}} \cr & {\text{6}}x - y = 25 \cr & \frac{{3x - y = - 5}}{{x\,\,\,\,\, = \,\,\,\,\,10}} \cr & \therefore y = 35 \cr & \frac{x}{y} = \frac{{10}}{{35}} = \frac{2}{7} \cr & {\text{Fraction = }}\frac{x}{y} = \frac{2}{7} \cr & {\text{Numerator = 2}} \cr & {\text{Denominator = 7}} \cr & {\text{LCM (Numerator}},{\text{Denominator}}) \cr & \Rightarrow 2 \times 7 = 14 \cr} $$

55.

If three number are 2a, 5a and 7a, what will be their LCM ?

• (A) 70a
• (B) 65a
• (C) 75a
• (D) 73a

Show Answer

 The given three numbers are 2a, 5a and 7a LCM of 2a, 5a and 7a = 2 × 5 × 7 × a = 70a

56.

The HCF and LCM of two numbers are 8 and 48 respectively. If one of the number is 24, then the other number is = ?

• (A) 48
• (B) 36
• (C) 24
• (D) 16

Show Answer

 HCF = 8 LCM = 48 One number = 24 Let other number be = y ∴ 24y = 48 × 8 ⇔ y = 16

57.

A milkman has 75 litres milk in one cane and 45 litres in another. The maximum capacity of container which can measure milk of either container exact number ?

• (A) 1 litre
• (B) 5 litres
• (C) 15 litres
• (D) 25 litres

Show Answer

 75 litres, 45 litres For maximum capacity take HCF (75, 45) = 15

58.

The HCF of two numbers is 29, and the other factors of their LCM are 15 and 13. The larger of the two numbers is:

• (A) 435
• (B) 377
• (C) 406
• (D) 464

Show Answer

 Let the numbers are a and b HCF = 29 LCM = HCF (a × b) LCM = 29(15 × 13) Largest number = 15 × 29 = 435

59.

The LCM of two numbers is 12 times their HCF. The sum of HCF and LCM is 403. If one number is 93, find the other = ?

• (A) 124
• (B) 128
• (C) 134
• (D) None of these

Show Answer

 $$\eqalign{ & {\text{Let HCF be }}h{\text{ and LCM be }}l \cr & {\text{Then }}l = {\text{ }}12h{\text{ and}} \cr & {\text{ }}l + h{\text{ }} = {\text{ }}403 \cr & \therefore 12h + h = 403{\text{ or }}h = {\text{ }}31 \cr & {\text{So }}l{\text{ = }}\left( {403 - 31} \right) = 372 \cr & {\text{Hence other number}} \cr & {\text{ = }}\left( {\frac{{31 \times 372}}{{93}}} \right) = 124 \cr} $$

60.

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

• (A) 276
• (B) 299
• (C) 322
• (D) 345

Show Answer

 The HCF of a group of numbers will be always a factor of their LCM. HCF is the product of all common prime factors using the least power of each common prime factor. LCM is the product of highest powers of all prime factors. Clearly, the numbers are (23 x 13) and (23 x 14) ∴ Larger number = (23 x 14) = 322